# How can $n!$ be represented in terms of $2^{n}$? [closed]

How can $n!$ be represented in terms of $2^{n}$?

I mean I want function $f$ such that $$n! = f(2^{n})$$

Also, I would like to know $$\frac{n!}{2^n} = g(n) = ?$$

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## closed as not a real question by Pete L. Clark, Ｊ. Ｍ., falagar, Rasmus, Robin ChapmanSep 5 '10 at 6:23

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

NOTE:question is edited :-) –  Pratik Deoghare Sep 2 '10 at 6:24
@Chandru1 No, not the limiting case. I want to know some $g(n)$. –  Pratik Deoghare Sep 2 '10 at 6:27
Here is a Kevin Buzzard-style remark: for your second question, the obvious answer is $g(n) = \frac{n!}{2^n}$. If you want any other answer, you need to explain why and in what way this is not a satisfactory answer for you. –  Pete L. Clark Sep 2 '10 at 7:12
@TheMachineCharmer: please stop and think for a moment to convince yourself that $g(n)$ cannot be given by a polynomial: it grows much more rapidly than $n^c$ for any constant $c$. I don't think there is any simpler exact expression. –  Pete L. Clark Sep 2 '10 at 7:55
I voted to reopen because the problem as it stands seems perfectly clear: writing x = 2^n yields n = lg(x), whence f(x) = lg(x)! (agreeing with Pete L. Clark's answer). The second, if interpreted in a similar vein (reading "g(x^n)" rather than "g(n)") would be answered by g(x) = lg(x)!/x. Whether either of these is interesting is another matter, but closing the question seems unnecessarily Draconian. –  whuber Sep 5 '10 at 14:23

Try $$f(t) = \Gamma ( 1 + \log_2 t ),$$ where $\log_2$ is the base 2 logarithm.